617. Merge Two Binary Trees

Given two binary trees and imagine that when you put one of them to cover the other, some nodes of the two trees are overlapped while the others are not.

You need to merge them into a new binary tree. The merge rule is that if two nodes overlap, then sum node values up as the new value of the merged node. Otherwise, the NOT null node will be used as the node of new tree.

Example 1:

Input: 
	Tree 1                     Tree 2                  
          1                         2                             
         / \                       / \                            
        3   2                     1   3                        
       /                           \   \                      
      5                             4   7                  
Output: 
Merged tree:
	     3
	    / \
	   4   5
	  / \   \ 
	 5   4   7

Note: The merging process must start from the root nodes of both trees

给定两颗二叉树，归并两棵树：将相同位置的结点的值加起来为新结点，并最终返回新的二叉树的根节点。

简单的思路是，从根节点开始，遍历两颗树的相同位置的每一个节点，判断树的结点是否为空，来决定是否要生成新结点以及确定结点的值。如果在该位置上，两个结点都为空，则不生成新结点；否则，生成新结点，并根据具体情况决定结点的值。利用递归：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) {
        if (!t1 && !t2)
            return nullptr;
        int value1 = t1 ? t1->val : 0;
        int value2 = t2 ? t2->val : 0;
        TreeNode* root = new TreeNode(value1 + value2);
        
        root->left = mergeTrees(t1 ? t1->left : nullptr, t2 ? t2->left : nullptr);
        root->right = mergeTrees(t1 ? t1->right : nullptr, t2 ? t2->right : nullptr);
        return root;
    }
};